A plano convex lens a thickness of 4 cm. Its radius of curvature is 20 cm, When its curved surface is kept on a horizontal surface and viewed from the top, its bottom appears to be raised by 1 cm. The focal length of the lens is

Options

A. 40 cm
B. 50 cm
C. 60 cm
D. 70 cm

Explanation

To find the focal length of the plano-convex lens, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a plano-convex lens with a thickness (t) of 4 cm and a radius of curvature (R) of 20 cm.
- The curved surface is placed on a horizontal surface, and when viewed from the top, it appears that the bottom of the lens is raised by 1 cm.

2. **Finding Actual and Apparent Depths:**
- The actual thickness of the lens is 4 cm.
- Since the bottom appears raised by 1 cm, the apparent depth (d_apparent) is:
\[ d_{apparent} = t - 1 \text{ cm} = 4 \text{ cm} - 1 \text{ cm} = 3 \text{ cm} \]

3. **Calculating the Refractive Index (μ):**
- The refractive index (μ) can be calculated using the formula:
\[ \mu = \frac{d_{actual}}{d_{apparent}} \]
- Substituting the values:
\[ \mu = \frac{4 \text{ cm}}{3 \text{ cm}} = \frac{4}{3} \]

4. **Using the Lens Maker's Formula:**
- The lens maker's formula is given by:
\[ \frac{1}{f} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For a plano-convex lens:
- \(R_1 = 20 \text{ cm}\) (curved surface)
- \(R_2 = \infty\) (plane surface, so \(\frac{1}{R_2} = 0\))
- Substituting the values into the formula:
\[ \frac{1}{f} = \left(\frac{4}{3} - 1\right) \left(\frac{1}{20} - 0\right) \]
- Simplifying:
\[ \frac{1}{f} = \frac{1}{3} \cdot \frac{1}{20} = \frac{1}{60} \]

5. **Finding the Focal Length (f):**
- Taking the reciprocal to find \(f\):
\[ f = 60 \text{ cm} \]

### Final Answer:
The focal length of the plano-convex lens is **60 cm**.

---